by Victor Bryant
Published Sunday March 06 2022
A group of us wanted to play a card game. We had a standard 52-card pack but we needed to deal out a whole pack with each person getting the same number of cards, so I added just enough “jokers” to make this possible. Then I randomly shuffled the enlarged pack (in this shuffled pack there was more than a 50:50 chance that there would be at least two jokers together). Then I dealt the cards out, each of us getting the same number (more than three). There was more than a 50:50 chance that my cards would not include a joker.
How many of us were there?
Sunday Times Teasers Discussion Group 
For a set with j jokers I calculated the propability for getting no joker in my c cards (c = factor of 52+j ).
But it seems to be not quite straightforward to calculate the probability of having at least two jokers together in the shuffled pack.
Bearing in mind that it is only necessary to find out if that probability is < or > 1/2 I found an answer for this Teaser, considering j in the range of 3 to 12.
Of the different card-packs that result in more than three cards to each player, only six cases yield a probability larger than 1/2 for jokers to be together. (Ignoring the minor anomalies for the top- and bottom card, facilitates the calculation). And among these cases only one has the required probability that a player’s hand has no jokers.
Interesting teaser. The author knows how many players there are, and adds the min no. of jokers to allow each player to be dealt the same number of cards. For any number of players this gives us a specific j number. I found two credible ways to do this, one of which agrees with Brian’s hidden result. Both have the same number of cards per player. So I have to assume that one is eliminated by the ‘two jokers together’ condition, but like Peter I found this difficult to calculate. But I’m not very good on permutations – it seems to me that the 2 adjacent jokers could be anywhere within the 52 card sequence, leading to some pretty big numbers? Maybe Erling has some additional knowledge, or have I misunderstood the text (again)?
I am not very good at probability calculations either but Jim Randell has put his method of calculating the ‘no two jokers in a row’ probability here and I get the same answer using a different approach as follows (where C(n, k) = n! / (k!(n-k)!) and ! is the factorial function):
To put J jokers in a pack of 52 cards with none adjacent we have to put them in the 51 gaps between the cards plus the two end positions – 53 in total. And the number of ways of placing J jokers in the 53 positions is C(53, J). The total number of ways to put J jokers in a pack of 52 cards is C(52+J, J) so the probability of no adjacent jokers in a random shuffle is C(53, J) / C(52 + J, J).
[Edited see below]
Thanks Brian
I did it much the same way – lay out 52+(j-1) cards, which leaves 52+(j-1) -1 gaps +2 ends = 52+j, and the remaining j can go in 2(j-1) of these.
On the second part, I found it easier to work out P for getting no jokers in a hand of h cards (h a factor of 52+j)
Brian
I think that your last sentence should begin: The total number of ways to put J jokers in a pack of 52 cards is C(52+J, J)
Thanks Tony, that would be better so I’ll change it.
Hi Robert
I calculated the probability in question analogous to the problem: If j friends randomly pick seats in a row with 52 + j seats; what are the chance no one get a seat next to a friend? First seat is ‘free’, but reduces free seats for the next pick with three seats (the one taken and its neighbours) and so on j times. The opposite outcome, that two or more are seated together is then of course 1 – p.
This approach ignores border-problems, but I take them to be neglectable in this case. (No big problem to include two exceptions though.)
The probabilities for adjacent jokers calculated my way, deviate from 0.5% to 3.6% compared to
P = C(53, j) / C(52 + j, j).
I thought this looked difficult but wasn’t too bad and I got the right answer. Having looked at the comments here though maybe I was fortunate that my simplistic approximation, or even wrong approach worked.
I took number of jokers N. For each joker in the pack there are N-1 out of remaining 51+N cards that may be next to it and 2 times for either side ignoring end of pack effects. So I had probability, approximate, of 2 jokers together 2N(N-1)/(51+N)… more later…
Setting my approximate probability of 2 jokers together gave me a lower limit on the number N of jokers.
I then wrote expressions for the probability of no jokers in a hand of C cards: =(52/(52+N))(51/(51+N)…) a multiplication of C terms, which is approx (52/(52+N))^C
This determines the maximum C for each N, the minimum C being 4. I then have a set of N and C values.
The number of people P=(52+N)/C
From the set of N and C there is only one case giving integer P, that P is the answer.
Brian’s method in thinking about the gaps is very helpful. Let the gaps between the cards and the two end spaces be called “slots”, with s slots, c cards and j jokers. s = c + 1.
The number of permutations (where order matters), for no more than one card in the same slot, is clearly s! / (s – j)!
Then thinking about the total number of permutations, it took me quite some time to see why it is not s^j. Put the first joker into a slot. Then the next joker can go into any of the slots, but in the slot which already has a card, the next card can go either side, and so there are effectively s + 1 slots. It turns out the addition of each card increases the effective number of slots by one. And so the total number of permutations is (s + j – 1)! / (s – 1)!, which, fortunately, agrees with Brian’s statement.
The probability of no jokers together = (53 / 53) * (52 / 54) * (51 / 55)….. with a total of j terms, and so the minimum number of jokers can be found very easily. Then it takes just three combinations of numbers of jokers and cards per player to find the answer and show that it is unique.
Thanks to Brian, I have learned things. And Victor Bryant gets more that a tip of the cap for a gem of a teaser.
As I noted earlier I obtained the answer without deriving the exact probability for having at least two jokers together.
But for curiosity I searched afterwards to find the exact value.
Rather than by clever thinking as done by John Crabtree (and probably by others) I obtained the number of arrangements with no jokers together “experimentaliy”, lining up j jokers with n other cards in different ways, like cases (n,j) = (3,2), (4,2), (4,3), (5,3). The probability follows with the overall number of arrangements C(n+j,j).